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My question refers to the paper http://arxiv.org/pdf/alg-geom/9710020.pdf where Batyrev proves that birational Calabi-Yau algebraic varieties have equal Betti numbers by counting points over finite fields using p-adic integration and so computes the Betti numbers using the Weil conjectures.

It seems that he is doing the following. Given a variety $X$ over $\mathbb{C}$, we can actually write it (and all the associated data we care about) as a variety $\mathcal{X}$ over $\mathcal{R}$, some finite-type $\mathbb{Z}$-algebra: i.e. such that $\mathcal{X} \otimes_\mathcal{R} \mathbb{C} = X$. We then fix an approriate maximal ideal $J(\pi)$ of $\mathcal{R}$ which lies above $p \in \mathbb{Z}$. I think we then turn our attention to the variety $\mathcal{X}\otimes_\mathcal{R} (\mathcal{R}/J(\pi))$, and using a ring of integers $R$ of a local number field with this as special fibre, we can count the number of points this variety has over every finite field extension of $\mathbb{F}_q = \mathcal{R}/J(\pi)$ using p-adic integration.

So the Weil conjectures give us the Betti numbers of this variety, and by proper smooth base change these Betti numbers are the same as those of $\mathcal{X} \otimes_\mathcal{R} \mathbb{C}$, but where the map $\mathcal{R} \rightarrow \mathbb{C}$ is not the natural inclusion but rather $\mathcal{R} \rightarrow R \hookrightarrow \mathbb{C}$. Since he is trying to compute the cohomology of the former, this doesn't make sense to me.

Can anyone see if I'm making a mistake somewhere? (or how my issue can be resolved?)

Thanks, Tom.

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  • $\begingroup$ Form a commutative square, which is hard to do in the comments. The bottom edge is $\mathcal{R}\subset \mathcal{K}$ where $\mathcal{K}$ is the fraction field; top $R\subset K$ also fraction field; left $\mathcal{R}\subset R$, and right $\mathcal{K}\subset K$. Now include $K\subset \mathbb{C}$. So all is well $\endgroup$
    – Donu Arapura
    Commented Mar 10, 2012 at 18:34
  • $\begingroup$ I should add that you should choose $K\subset \mathbb{C}$ to be compatible with $\mathcal{K}\subset \mathbb{C}$. I leave it to you to see that this is possible. $\endgroup$
    – Donu Arapura
    Commented Mar 10, 2012 at 18:44
  • $\begingroup$ I don't think there's any kind of natural embedding of $\mathcal{R}$ into $R$ :the paper seems to go via $\mathcal{R} \rightarrow \mathcal{R}\otimes_\mathbb{Z} \mathbb{Z}_p \rightarrow R$ which feels unlikely to be injective. For example, if $\mathcal{R}$ is something like $\mathbb{Z}[x_1,...,x_n]$. So do you mean I should artificially construct some kind of injection $\mathcal{R} \hookrightarrow R$,using the fact $\mathbb{Z}_p$ has lots of transcendental elements,and then play around to make it compatible with the inclusion of $\mathcal{R}$ in $\mathbb{C}$? $\endgroup$
    – Tom Lovering
    Commented Mar 12, 2012 at 12:43
  • $\begingroup$ Unless I misunderstood your notation, $\mathcal{R}$ is an integral domain. You first localize $\mathcal{R}\to \mathcal{R}_{J(\pi)}=S$ and then complete to get $S\to R=\hat S$. Both maps are injective. The unnatural embedding is $R\to \mathbb{C}$, and here you do have to play around. If you prefer, another way to do this is to observe that the Betti numbers are the same for any two complex embeddings, because they both coincide with dimension of etale cohomology groups. $\endgroup$
    – Donu Arapura
    Commented Mar 12, 2012 at 13:30
  • $\begingroup$ Thanks for the quick reply. I think it's helped focus in on my confusion. If $R$ were constructed as you say, I'm suspicious that its function field might have some transcendence degree over $\mathbb{Q}_p$. On the other hand, the paper seems to say that $R$ is "the maximal compact subring in a local $p$-adic field," which I took to mean an algebraic extension of $\mathbb{Q}_p$, and I think actually being a local field is necessary to make the rest of the argument work. So I guess my question now is: is $Frac(R)$ (as you define it) obviously algebraic $/\mathbb{Q}_p$? $\endgroup$
    – Tom Lovering
    Commented Mar 12, 2012 at 14:22

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